Difference between revisions of "2006 AMC 10B Problems/Problem 17"

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== Solution ==
 
== Solution ==
Since there is the same ammounts of all the balls in Alice's bag and Bob's bag, and there is an equal chance of each ball being selected, the color of the ball that Alice puts in Bob's bag doesn't matter. [[Without loss of generality]], let the ball Alice puts in Bob's bag be red.  
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Since there is the same amounts of all the balls in Alice's bag and Bob's bag, and there is an equal chance of each ball being selected, the color of the ball that Alice puts in Bob's bag doesn't matter. [[Without loss of generality]], let the ball Alice puts in Bob's bag be red.  
  
 
For both bags to have the same contents, Bob must select one of the 2 red balls out of the 6 balls in his bag.  
 
For both bags to have the same contents, Bob must select one of the 2 red balls out of the 6 balls in his bag.  

Revision as of 12:14, 18 July 2006

Problem

Bob and Alice each have a bag that contains one ball of each of the colors blue, green, orange, red, and violet. Alice randomly selects one ball from her bag and puts it into Bob's bag. Bob then randomly selects one ball from his bag and puts it into Alice's bag. What is the probability that after this process the contents of the two bags are the same?

$\mathrm{(A) \ } \frac{1}{10}\qquad \mathrm{(B) \ } \frac{1}{6}\qquad \mathrm{(C) \ } \frac{1}{5}\qquad \mathrm{(D) \ } \frac{1}{3}\qquad \mathrm{(E) \ } \frac{1}{2}$

Solution

Since there is the same amounts of all the balls in Alice's bag and Bob's bag, and there is an equal chance of each ball being selected, the color of the ball that Alice puts in Bob's bag doesn't matter. Without loss of generality, let the ball Alice puts in Bob's bag be red.

For both bags to have the same contents, Bob must select one of the 2 red balls out of the 6 balls in his bag.

So the desired probability is $\frac{2}{6} = \frac{1}{3} \Rightarrow D$

See Also